游戏人生Silverlight(4) - 连连看[Silverlight 2.0(c#)]

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游戏人生Silverlight(4) - 连连看[Silverlight 2.0(c#)]

作者：webabcd


介绍
使用 Silverlight 2.0(c#) 开发一个连连看游戏


玩法
用鼠标左键选中卡片，如果选中的两卡片间的连线不多于 3 根直线，则选中的两卡片可消除


在线DEMO



思路
1、卡片初始排列算法：已知容器容量为 x, 不重复的卡片数量为 y, x >= y && x % 2 == 0, 首先在容器内随机排列卡片，然后取出容器内相同的卡片个数为奇数的集合（集合内成员数量必为偶数个），最后将该集合一刀切，将集合右半部分的卡片的依次复制到集合左半部分。以上算法保证了在一定随机率的基础上，不会出现相同的卡片个数为奇数的情况
2、无解算法和重排算法：在容器内存在的卡片中，两两计算是否存在可消路径，如果没有就是无解，需要重排。重排时，需要得到现存的卡片集合和卡片位置集合，在卡片集合中随机取卡片（取出一个，原集合就要移除这一个），然后依次放到卡片位置集合内，从而达到将现存卡片重新排列的目的
3、两点消去路径的算法以及取最优消去路径的算法：取玩家选的第一点的 x 轴方向和 y 轴方向上的所有无占位符的坐标集合（包括自己），名称分别为 x1s, y1s；取玩家选的第二点的 x 轴方向和 y 轴方向上的所有无占位符的坐标集合（包括自己），名称分别为 x2s, y2s。先在 x1s 和 x2s 中找 x 坐标相等的两点，然后找出该两点与玩家选的两点可组成一条连续的直线的集合，该集合就是可消路径的集合，之后同理再在 y1s 和 y2s 中找到可消路径的集合。两集合合并就是玩家选中的两点间的所有可消路径的集合，该集合为空则两点不可消，该集合内的最短路径则为最优消去路径，集合内的 4 点连接线则为消去路径的连接线
4、游戏使用MVVM(Model - View - ViewModel)模式开发


关键代码
Core.cs

using System;
using System.Net;
using System.Windows;
using System.Windows.Controls;
using System.Windows.Documents;
using System.Windows.Ink;
using System.Windows.Input;
using System.Windows.Media;
using System.Windows.Media.Animation;
using System.Windows.Shapes;

using YYMatch.Models;
using System.Collections.ObjectModel;
using System.Linq;
using System.Collections.Generic;
using System.ComponentModel;
using System.Threading;

namespace YYMatch.ViewModels
{
         /// <summary>
         /// 连连看核心模块
         /// </summary>
         public class Core : INotifyPropertyChanged
        {
                ObservableCollection<CardModel> _cards = null;
                 int _rows = 0;
                 int _columns = 0;
                SynchronizationContext _syncContext = null;

                 public Core()
                {
                         // 在容器上布满空的卡片
                        _cards = new ObservableCollection<CardModel>();
                         for ( int i = 0; i < Global.ContainerColumns * Global.ContainerRows; i++)
                        {
                                _cards.Add( new CardModel( "00", i));
                        }

                        _syncContext = SynchronizationContext.Current;
                }

                 public void Start( int rows, int columns)
                {
                        _rows = rows;
                        _columns = columns;

                        InitCard();
                }

                 private ObservableCollection<CardModel> InitCard()
                {
                        Random r = new Random();
                         // 卡片集合在容器内的范围
                         int minX = (Global.ContainerColumns - _columns) / 2;
                         int maxX = minX + _columns - 1;
                         int minY = (Global.ContainerRows - _rows) / 2;
                         int maxY = minY + _rows - 1;

                         for ( int x = 0; x < Global.ContainerColumns; x++)
                        {
                                 for ( int y = 0; y < Global.ContainerRows; y++)
                                {
                                         // 18 张图随机排列
                                         string imageName = r.Next(1, Global.ImageCount + 1).ToString().PadLeft(2, '0');
                                        var cardPoint = new CardPoint(x, y);

                                         if (x >= minX && x <= maxX && y >= minY && y <= maxY)
                                        {
                                                _cards[cardPoint.Position] = new CardModel(imageName, cardPoint.Position);
                                        }
                                }
                        }

                         // 相同的卡片个数为奇数的集合
                        var oddImages = _cards.Where(p => p.ImageName != Global.EmptyImageName)
                                .GroupBy(p => p.ImageName)
                                .Select(p => new { ImageName = p.Key, Count = p.Count() })
                                .Where(p => p.Count % 2 > 0)
                                .ToList();

                         // 如果 oddImages 集合的成员为奇数个（保证容器容量为偶数个则不可能出现这种情况）
                         if (oddImages.Count() % 2 > 0)
                        {
                                 throw new Exception( "无法初始化程序");
                        }
                         else
                        {
                                 // 在集合中将所有的个数为奇数的卡片各自取出一个放到 temp 中
                                 // 将 temp 一刀切，使其右半部分的卡片的 ImageName 依次赋值为左半部分的卡片的 ImageName
                                 // 由此保证相同的卡片均为偶数个
                                List<CardModel> tempCards = new List<CardModel>();
                                 for ( int i = 0; i < oddImages.Count(); i++)
                                {
                                         if (i < oddImages.Count() / 2)
                                        {
                                                var tempCard = _cards.Last(p => p.ImageName == oddImages.ElementAt(i).ImageName);
                                                tempCards.Add(tempCard);
                                        }
                                         else
                                        {
                                                var tempCard = _cards.Last(p => p.ImageName == oddImages.ElementAt(i).ImageName);
                                                _cards[tempCard.Position].ImageName = tempCards[i - oddImages.Count() / 2].ImageName;
                                        }
                                }
                        }

                         if (!IsActive())
                                Replace();

                         return _cards;
                }

                 /// <summary>
                 /// 判断两卡片是否可消
                 /// </summary>
                 public bool Match(CardModel c1, CardModel c2, bool removeCard)
                {
                         bool result = false;

                         if (c1.ImageName != c2.ImageName
                                || c1.ImageName == Global.EmptyImageName
                                || c2.ImageName == Global.EmptyImageName
                                || c1.Position == c2.Position)
                                 return false;

                         // 如果可消的话，则 point1, point2, point3, point4 会组成消去两卡片的路径（共4个点）
                        CardPoint point1 = new CardPoint(0);
                        CardPoint point2 = new CardPoint(0);
                        CardPoint point3 = new CardPoint(0);
                        CardPoint point4 = new CardPoint(0);
                         // 最小路径长度
                         int minLength = int.MaxValue;

                        CardPoint p1 = new CardPoint(c1.Position);
                        CardPoint p2 = new CardPoint(c2.Position);

                        var p1xs = GetXPositions(p1);
                        var p1ys = GetYPositions(p1);
                        var p2xs = GetXPositions(p2);
                        var p2ys = GetYPositions(p2);

                         // 在两点各自的 X 轴方向上找可消点（两个可消点的 X 坐标相等）
                        var pxs = from p1x in p1xs
                                            join p2x in p2xs
                                            on p1x.X equals p2x.X
                                            select new { p1x, p2x };
                         foreach (var px in pxs)
                        {
                                 if (MatchLine(p1, px.p1x) && MatchLine(px.p1x, px.p2x) && MatchLine(px.p2x, p2))
                                {
                                         int length = Math.Abs(p1.X - px.p1x.X) + Math.Abs(px.p1x.Y - px.p2x.Y) + Math.Abs(px.p2x.X - p2.X);

                                         // 查找最短连接路径
                                         if (length < minLength)
                                        {
                                                minLength = length;
                                                point1 = p1;
                                                point2 = px.p1x;
                                                point3 = px.p2x;
                                                point4 = p2;
                                        }
                                        result = true;
                                }
                        }

                         // 在两点各自的 Y 轴方向上找可消点（两个可消点的 Y 坐标相等）
                        var pys = from p1y in p1ys
                                            join p2y in p2ys
                                            on p1y.Y equals p2y.Y
                                            select new { p1y, p2y };
                         foreach (var py in pys)
                        {
                                 if (MatchLine(p1, py.p1y) && MatchLine(py.p1y, py.p2y) && MatchLine(py.p2y, p2))
                                {
                                         int length = Math.Abs(p1.Y - py.p1y.Y) + Math.Abs(py.p1y.X - py.p2y.X) + Math.Abs(py.p2y.Y - p2.Y);

                                         // 查找最短连接路径
                                         if (length < minLength)
                                        {
                                                minLength = length;
                                                point1 = p1;
                                                point2 = py.p1y;
                                                point3 = py.p2y;
                                                point4 = p2;
                                        }
                                        result = true;
                                }
                        }

                         if (removeCard && result)
                        {
                                RemoveCard(c1, c2, point1, point2, point3, point4);
                        }

                         return result;
                }

                 /// <summary>
                 /// 直线上的两个 CardPoint 是否可以消去
                 /// </summary>
                 private bool MatchLine(CardPoint p1, CardPoint p2)
                {
                         if (p1.X != p2.X && p2.Y != p2.Y)
                                 return false;

                        var range = _cards.Where(p =>
                                p.Position > Math.Min(p1.Position, p2.Position)
                                && p.Position < Math.Max(p1.Position, p2.Position));

                         if (p1.X == p2.X)
                        {
                                range = range.Where(p => (p.Position - p1.Position) % Global.ContainerColumns == 0);
                        };

                         if (range.Count() == 0 || range.All(p => p.ImageName == Global.EmptyImageName))
                                 return true;

                         return false;
                }

                 /// <summary>
                 /// 获取指定的 CardPoint 的 X 轴方向上的所有 ImageName 为 Global.EmptyImageName 的 CardPoint 集合
                 /// </summary>
                 private List<CardPoint> GetXPositions(CardPoint p)
                {
                        var result = new List<CardPoint>() { p };
                         for ( int i = 0; i < Global.ContainerColumns; i++)
                        {
                                var point = new CardPoint(p.Y * Global.ContainerColumns + i);
                                 if (_cards[point.Position].ImageName == Global.EmptyImageName)
                                        result.Add(point);
                        }

                         return result;
                }

 

未完待续>>